geometry help!

[iluvchairs112]

i'm wondering if anyone can help me with this geometry question ... ok if you are given 2 points in R3, how would you go about finding the plane where all the points on the plane are equal distances from both points?

[octopi]

You should be able to find a line(vector) that is perpendicular to the line between a,b, that passes through the midpoint.
Then you find another line(vector) that is perpendicular (they're should be many) you'd use the dot product to find this.
Then you use the two line(vectors) you've found to form the plane.

[Cervantes]

octopi @ Sun Apr 15, 2007 5:28 pm wrote:

You should be able to find a line(vector) that is perpendicular to the line between a,b, that passes through the midpoint.
Then you find another line(vector) that is perpendicular (they're should be many) you'd use the dot product to find this.
Then you use the two line(vectors) you've found to form the plane.

You'd use the cross product.

[klopyrev]

You need a plane that is perpendicular to the line between the two points and that passes through their midpoint. A plane is defined by an equation of the form Ax+By+Cz = D. If I remember correctly, (A,B,C) is the normal vector. It is perpendicular to the plane. In your case, (A,B,C) would be the vector between the two points, since you want the plane perpendicular to that vector. So, to find the plane between points P(a,b,c) and Q(d,e,f), you find the equation of the plane with A = (a-d), B = (b-e) and C = (c-f). So, (a-d)x+(b-e)y+(c-f)z = D. Sub in ((a+d)/2, (b+e)/2, (c+f)/2) to get the value for D and you are done. Hope that helps.

KL

[Brightguy]

If n is the direction of the line passing through the points and m is the midpoint of the points, then n is a normal vector to the plane and m is a point on the plane.

Consequently, the equation of the plane is given by . Or you can just remember the equation will have form ax+by+cz+d=0, once you know n=(a,b,c) and a point on the plane you can solve for d.